06-240/Term Test

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The Test

Front Page

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Math 240 Algebra I - Term Test

University of Toronto, October 24, 2006

Solve the 5 problems on the other side of this page.

Each of the problems is worth 20 points.

You have an hour and 45 minutes.

Notes.

• No outside material other than stationary and a basic calculator is allowed.
• We will have an extra hour of class time in our regular class room on Thursday, replacing the first tutorial hour.
• The final exam date was posted by the faculty - it will take place on Wednesday December 13 from 2PM until 5PM at room 3 of the Clara Benson Building, 320 Huron Street (south west of Harbord cross Huron, home of the Faculty of Physical Education and Health).

Good Luck!

Questions Page

Solve the following 5 problems. Each of the problems is worth 20 points. You have an hour and 45 minutes.

Problem 1. Let $F$ be a field with zero element $0_{F}$, let $V$ be a vector space with zero element $0_{V}$ and let $v\in V$ be some vector. Using only the axioms of fields and vector spaces, prove that $0_{F}\cdot v=0_{V}$.

Problem 2.

1. In the field ${\mathbb {C} }$ of complex numbers, compute ${\frac {1}{2+3i}}+{\frac {1}{2-3i}}$      and     ${\frac {1}{2+3i}}-{\frac {1}{2-3i}}$.
2. Working in the field ${\mathbb {Z} }/7$ of integers modulo 7, make a table showing the values of $a^{-1}$ for every $a\neq 0$.

Problem 3. Let $V$ be a vector space and let $W_{1}$ and $W_{2}$ be subspaces of $V$. Prove that $W_{1}\cup W_{2}$ is a subspace of $V$ iff $W_{1}\subset W_{2}$ or $W_{2}\subset W_{1}$.

Problem 4. In the vector space $M_{2\times 2}({\mathbb {Q} })$, decide if the matrix ${\begin{pmatrix}1&2\\-3&4\end{pmatrix}}$ is a linear combination of the elements of $S=\left\{{\begin{pmatrix}1&0\\-1&0\end{pmatrix}},\ {\begin{pmatrix}0&1\\0&1\end{pmatrix}},\ {\begin{pmatrix}1&1\\0&0\end{pmatrix}}\right\}$.

Problem 5. Let $V$ be a finite dimensional vector space and let $W_{1}$ and $W_{2}$ be subspaces of $V$ for which $W_{1}\cap W_{2}=\{0\}$. Denote the linear span of $W_{1}\cup W_{2}$ by $W_{1}+W_{2}$. Prove that $\dim(W_{1}+W_{2})=\dim W_{1}+\dim W_{2}$.

Good Luck!

[print]

The Results

Excluding a few late exams and before appeals, 63 students took the exam; The average grade was 67.46 and the standard deviation was 24.52.

The results are quite similar to what I expected them to be. The easiest questions (on average) were the computational ones, the hardest were the ones involving proofs.

How should you read your grade?

• If you got 100 you should pat yourself on your shoulder and feel good.
• If you got something like 95, you're doing great. You made a few relatively minor mistakes; find out what they are and try to avoid them next time.
• If you got something like 80 you're doing fine but you did miss something significant, probably more than just a minor thing. Figure out what it was and make a plan to fix the problem for next time.
• If you got something like 60 you should be concerned. You are still in position to improve greatly and get an excellent grade at the end, but what you missed is quite significant and you are at the risk of finding yourself far behind. You must analyze what happened - perhaps it was a minor mishap, but more likely you misunderstood something major or something major is missing in your background. Find out what it is and try to come up with a realistic strategy to overcome the difficulty!
• If you got something like 35, most likely you are not gaining much from this class and you should consider dropping it, unless you are convinced that you fully understand the cause of your difficulty (you were very sick, you really couldn't study at all for the two weeks before the exam because of some unusual circumstances, something like that) and you feel confident you have a fix for next time. If you do decide to drop the class, don't feel too bad about it. It is the hardest first year algebra class at UofT and of the thousands of students taking math here, very few come with sufficient preparation to do well in it.

Note that problems with writing are problems, period. Perhaps you got a low grade but you feel you know the material enough for a high grade only you didn't write everything you know or you didn't it write well enough or the silly graders simply didn't get what you wrote (and it isn't a simple misunderstanding - see "appeals" below). If this describes you, don't underestimate your problem. If you don't process and resolve it, it is likely to recur.

Appeals.

Remember! Grading is a difficult process and mistakes always happen - solutions get misread, parts are forgotten, grades are not added up correctly. You must read your exam and make sure that you understand how it was graded. If you disagree with anything, don't hesitate to complain! Your first stop should be the person who graded the problem in question, and only if you can't agree with him you should appeal to Dror.

Problem 1 and 2 were graded by Dmitry Donin, problem 3 was graded by Dror and problems 4 and 5 were graded by Paul Lee.

The deadline to start the appeal process is Thursday November 2 at 4PM.

Solution Set

Students are most welcome to post a solution set here.

WARNING: The solution set below, written for students and by students, is provided "as is", with absolutely no warranty. It can not be assumed to be complete, correct, reliable or relevant. If you don't like it, don't read it. Visit this pages' history tab to see who added what and when.

Problem 1. $0_{F}\cdot v=(0_{F}+0_{F})\cdot v$ (by F3)

$(0_{F}+0_{F})\cdot v=0_{F}\cdot v+0_{F}\cdot v$ (by VS8)

By VS4, $\exists \ (0_{F}\cdot v)'s.t.(0_{F}\cdot v)+(0_{F}\cdot v)'=0_{V}$

Add $(0_{F}\cdot v)'$ to both sides of $0_{F}\cdot v=0_{F}\cdot v+0_{F}\cdot v$

$(0_{F}\cdot v)'+(0_{F}\cdot v)=[(0_{F}\cdot v)'+0_{F}\cdot v]+0_{F}\cdot v$

$0_{V}=0_{V}+0_{F}\cdot v$ (by construction)

$0_{V}=0_{F}\cdot v$ (by VS3)

Problem 2. 1) ${\frac {1}{2+3i}}+{\frac {1}{2-3i}}={\frac {(2-3i)+(2+3i)}{(2+3i)(2-3i)}}={\frac {4}{2^{2}+3^{3}}}={\frac {4}{13}}$

${\frac {1}{2+3i}}-{\frac {1}{2-3i}}={\frac {(2-3i)-(2+3i)}{(2+3i)(2-3i)}}={\frac {-6i}{2^{2}+3^{3}}}=-{\frac {6}{13}}i$

2) $1^{-1}=1$

$2^{-1}=4$

$3^{-1}=5$

$4^{-1}=2$

$5^{-1}=3$

$6^{-1}=6$

Problem 3. First suppose $W_{1}\subset W_{2}$ or $W_{2}\subset W_{1}$, then $W_{1}\cup W_{2}$ equals $W_{2}$ or $W_{1}$. Either case $W_{1}\cup W_{2}$ is a subspace of $V$.

Now suppose $W_{1}\cup W_{2}$ is a subspace of $V$ and neither $W_{1}\subset W_{2}$ nor $W_{2}\subset W_{1}$. It must follow that $\exists \ x\in W_{1}$ s.t. $x$ is not in $W_{2}$ and $\exists \ y\in W_{2}$ s.t. $y$ is not in $W_{1}$. Since $x,y\in W_{1}\cup W_{2}$, $x+y\in W_{1}\cup W_{2}$. If $x+y\in W_{1}$, then $\exists \ -x\in W_{1}$ s.t. $(-x)+x+y\in W_{1}$ and $(-x)+x=0$, so $y\in W_{1}$, a contradiction. If $x+y\in W_{2}$, then $\exists \ -y\in W_{2}$ s.t. $x+y+(-y)\in W_{2}$ and $(-y)+y=0$, so $x\in W_{2}$, a contradiction. Therefore either $W_{1}\subset W_{2}$ or $W_{2}\subset W_{1}$.

Problem 4. Suppose ${\begin{pmatrix}1&2\\-3&4\end{pmatrix}}$ is a linear combination of ${\begin{pmatrix}1&0\\-1&0\end{pmatrix}}$, ${\begin{pmatrix}0&1\\0&1\end{pmatrix}}$, and ${\begin{pmatrix}1&1\\0&0\end{pmatrix}}$, then $\exists \ a_{1},a_{2},a_{3}\in F$ s.t. ${\begin{pmatrix}1&2\\-3&4\end{pmatrix}}=a_{1}{\begin{pmatrix}1&0\\-1&0\end{pmatrix}}+a_{2}{\begin{pmatrix}0&1\\0&1\end{pmatrix}}+a_{3}{\begin{pmatrix}1&1\\0&0\end{pmatrix}}$.

${\mbox{Need to solve}}{\begin{cases}1=a_{1}^{}+a_{3}^{}\\2=a_{2}^{}+a_{3}^{}\\-3=-a_{1}^{}\\4=a_{2}^{}\end{cases}}$

Solving the equations yields $a_{1}=3$, $a_{2}=4$, $a_{3}=-2$, so ${\begin{pmatrix}1&2\\-3&4\end{pmatrix}}$ is a linear combination of ${\begin{pmatrix}1&0\\-1&0\end{pmatrix}}$, ${\begin{pmatrix}0&1\\0&1\end{pmatrix}}$, and ${\begin{pmatrix}1&1\\0&0\end{pmatrix}}$. Specifically, ${\begin{pmatrix}1&2\\-3&4\end{pmatrix}}=3{\begin{pmatrix}1&0\\-1&0\end{pmatrix}}+4{\begin{pmatrix}0&1\\0&1\end{pmatrix}}-2{\begin{pmatrix}1&1\\0&0\end{pmatrix}}$.

Problem 5. Since $V$ is finite-dimensional, then so are $W_{1}$ and $W_{2}$ and their basis. Let ${a_{1},a_{2},...,a_{m}}$ be the basis of $W_{1}$ and ${b_{1},b_{2},...,b_{n}}$ be the basis of $W_{2}$ so $\dim W_{1}=m$ and $\dim W_{2}=n$.

We know ${a_{1},a_{2},...,a_{m}}$ and ${b_{1},b_{2},...,b_{n}}$ are linearly independent and clearly ${a_{1},a_{2},...,a_{m},b_{1},b_{2},...,b_{n}}$ spans $W_{1}+W_{2}$. If ${a_{1},a_{2},...,a_{m},b_{1},b_{2},...,b_{n}}$ is linearly dependent, then there exist not all zero coefficients $c_{1},c_{2},...,c_{m+n}\in F$ s.t. $c_{1}a_{1}+c_{2}a_{2}+...+c_{m}a_{m}+c_{m+1}b_{1}+c_{m+2}b_{2}+...+c_{m+n}b_{n}=0$. Then some linear combinations of ${a_{1},a_{2},...,a_{m}}$ with not all zero coefficients can be expressed in linear combinations of ${b_{1},b_{2},...,b_{n}}$, but this would imply $W_{1}\cap W_{2}\neq \{0\}$, a contradiction. Therefore ${a_{1},a_{2},...,a_{m},b_{1},b_{2},...,b_{n}}$ is a linearly independent set that spans $W_{1}+W_{2}$, it's the basis of $W_{1}+W_{2}$. We have $\dim(W_{1}+W_{2})=m+n=\dim W_{1}+\dim W_{2}$.